Problem: Simplify and expand the following expression: $ \dfrac{3}{5r + 50}+ \dfrac{1}{3r - 21}- \dfrac{r}{r^2 + 3r - 70} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{3}{5r + 50} = \dfrac{3}{5(r + 10)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{1}{3r - 21} = \dfrac{1}{3(r - 7)}$ We can factor the quadratic in the third term: $ \dfrac{r}{r^2 + 3r - 70} = \dfrac{r}{(r + 10)(r - 7)}$ Now we have: $ \dfrac{3}{5(r + 10)}+ \dfrac{1}{3(r - 7)}- \dfrac{r}{(r + 10)(r - 7)} $ The least common multiple of the denominators is: $ 15(r + 10)(r - 7)$ In order to get the first term over $15(r + 10)(r - 7)$ , multiply by $\dfrac{3(r - 7)}{3(r - 7)}$ $ \dfrac{3}{5(r + 10)} \times \dfrac{3(r - 7)}{3(r - 7)} = \dfrac{9(r - 7)}{15(r + 10)(r - 7)} $ In order to get the second term over $15(r + 10)(r - 7)$ , multiply by $\dfrac{5(r + 10)}{5(r + 10)}$ $ \dfrac{1}{3(r - 7)} \times \dfrac{5(r + 10)}{5(r + 10)} = \dfrac{5(r + 10)}{15(r + 10)(r - 7)} $ In order to get the third term over $15(r + 10)(r - 7)$ , multiply by $\dfrac{15}{15}$ $ \dfrac{r}{(r + 10)(r - 7)} \times \dfrac{15}{15} = \dfrac{15r}{15(r + 10)(r - 7)} $ Now we have: $ \dfrac{9(r - 7)}{15(r + 10)(r - 7)} + \dfrac{5(r + 10)}{15(r + 10)(r - 7)} - \dfrac{15r}{15(r + 10)(r - 7)} $ $ = \dfrac{ 9(r - 7) + 5(r + 10) - 15r} {15(r + 10)(r - 7)} $ Expand: $ = \dfrac{9r - 63 + 5r + 50 - 15r}{15r^2 + 45r - 1050} $ $ = \dfrac{-r - 13}{15r^2 + 45r - 1050}$